10 Two Sum

Key Pair | Practice | GeeksforGeeks

Question

Given an array Arr of N positive integers and another number X. Determine whether or not there exist two elements in Arr whose sum is exactly X.

Example 1:

Input:
N = 6, X = 16
Arr[] = {1, 4, 45, 6, 10, 8}
Output: Yes
Explanation: Arr[3] + Arr[4] = 6 + 10 = 16

Example 2:

Input:
N = 5, X = 10
Arr[] = {1, 2, 4, 3, 6}
Output: Yes
Explanation: Arr[2] + Arr[4] = 4 + 6 = 10

Your Task: You don't need to read input or print anything. Your task is to complete the function hasArrayTwoCandidates() which takes the array of integers arr, n and x as parameters and returns boolean denoting the answer.

Expected Time Complexity: O(N) Expected Auxiliary Space: O(N)

Constraints: 1 ≤ N ≤ 105 1 ≤ Arr[i] ≤ 105

Naive Approach

Logic

Simply use two loops and check for every pair

Code

bool hasArrayTwoCandidates(int arr[], int n, int x) {
    for(int i = 0; i < n; i++)
        for(int j = i + 1; j < n; j++) 
            if(arr[i] + arr[j] == x) 
                return true;
    return false;
}

Complexity

Time Compleixty: $O(n^2)$

Space Complexity: $O(1)$

Using `map`

Logic

1. For each element(say a) there must be a b such that a + b = x

2. Traverse the given array as a

   1. Since a + b = x

   2. b = x - a

   3. Check the map to see if b has already been found

   4. If found, return true

      1. As we have proven that the a and b necessary to make a + b = x are present in the array

   5. if not, add a to the map

      1. This will be used as b to make pair with some other a for next iterations.

Code

bool hasArrayTwoCandidates(int arr[], int n, int x) {
    unordered_set<int> visited;
    
    for(int i = 0; i < n; i++) 
        if(visited.count(x - arr[i]))
            return true;
        else 
            visited.insert(arr[i]);
            
   return false;
}

Complexity

Time Complexity: $O(n)$

Space Complexity: $O(n)$